Why do rockets have multiple stages? What is the advantage for rockets to have multiple stages? 
Wouldn't a single stage with the same amount of fuel weigh less?
Note I would like a quantitative answer, if possible :-)

 A: Launch weight would be lower if you had a fixed fuel load, and only one engine and fuel tank. However the specific impulse applied to the payload would be lower. The problem is that even when the fuel is say 90% exhausted, the rocket is still trying to accelerate the now grossly oversized fuel tank and engine. So the trick is to try to reduce the deadweight (structural mass) as the fuel is consumed. Another compromise system is to have jetisonable external fuel tanks, like the shuttle, which are thrown away once their fuel is consumed.
A: As Omega Centauri wrote, it is mainly about removing unused tank mass; for numbers, see Wikipedia article about Ciołkowski's equation, especially the example there.
A: The easiest way to think of it is this, imagine all the mass left over when a rocket has burned 85% of it's fuel. The mass of most of the tank and structure is now overkill and waste. It would be nice to be able to jettison that extra mass so that the fuel left can accelerate only the payload.
That's what a multi-stage rocket does. It jettisons the mass of initial stages so that the remaining fuel and thrust can accelerate much smaller mass to a much higher velocity than it would have been able to if there was only one stage. Remember acceleration is proportional to mass, so if you can get rid of say 80% of the mass then you can accelerate  the payload 5 times more for the same remaining fuel.
Another benefit is that you can use rocket motors that are tuned for different velocities.  In the initial stage you need maximum thrust and the rocket is not moving as fast. In the later stages you want high efficiency motors, not necessarily high thrust.
To get very high velocities it requires less overall fuel and mass with multiple stages. This comes at the cost of greater complexity and cost.
A: Another aspect to consider is the burn characteristics of the rocket motors. This is especially important in solid rocket motors because once lit, they are self-oxidizing and are not easy to turn off. 
At low altitudes, the rocket should not accelerate too rapidly because the air is very dense and the power required is proportional to velocity cubed. So you just want to get the thing going until you reach an altitude where the air is less dense and it's more economical to go fast. So you may have a first stage that burns relatively slowly. 
Once that burns and you reach a higher altitude where you can go faster, you drop your "slow burn" motor and kick on your powerful motor. Now the air density is much lower so you can accelerate as quickly as you want and reach whatever speed is needed. 
A third stage might be used to fine-tune the speed and position the craft in whatever orbit or trajectory is needed. 
Furthermore, because the air pressure decreases with altitude, the ideal nozzle shape is not the same at low altitude as high. An overexpanded nozzle at sea-level can be an underexpanded nozzle at altitude resulting in a very narrow range where it is operating at maximum efficiency. 
You could design adaptive nozzles but they are very heavy and expensive, and they can't really be made to service the full range. Or, you could have stages with fixed nozzles that are designed to be as efficient as possible over the range of altitudes serviced. 
So, in addition to the answers above about dropping weight as you go resulting in less fuel use, each stage can also be designed to take into account the operating regime and needs by carefully selecting the fuel for appropriate thrust/burn rates and by designing the nozzle for the nominal operating conditions for the altitudes serviced by the motor. Both of which lead to a much more efficient motor and thus less fuel.
A: Edited a little now that I better understand your question.
Short
In a multi-stage, the weight of the parts that are dropped along the ride compensates for the fact that the extra engines make it heavier in the beginning. Partially because a rocket's engine isn't that heavy compared to the fuel tank. The engine mostly just ignites and controls the combustion, while the fuel tank needs to be huge.

Long
I'll analyse the very optimal case, in which the energy spent is the minimum energy needed to get your rocket in orbit. Real cases go waaay above this and it's not even practical to do energy balance. Thornton & Marion does a great (undergraduate level) analysis using linear momentum.
In order to launch a single stage rocket, there's a specific size that minimizes the needed energy. The weight of your rocket (just the hull, not the fuel) increases with size $P_R=g\rho V$. And the minimum amount of energy necessary to get your rocket in orbit obviously increases with its weight: $$W_{min} = hP_R + W',$$ where $W'$ is the work you do to lift the fuel. $W'$ is not important here but, for the record, it goes something like:
$$W' = g\int_0^h m_{fuel}(z)dz'.$$
So, if the rocket is too small, the little amount of fuel you can fit into it does not contain enough energy to lift the hull. If the rocket is too big, then you're just wasting energy because it's unnecessarily heavy. That means you have an optimal size which minimizes the energy of a single-stage rocket. Once you find it, the amount of fuel associated to that size is the very minimum you'll need to get your rocket on orbit.
Now consider a rocket of that same size, but with 2 stages. Let's assume that it has an extra engine of weight $p_e$. Let's also assume that, when your rocket reaches a height $h/2$ more than $1/3$ of its tank will be empty (which is true). Since that part of my tank is empty, I might as well leave it behind (along with the engine) and drop the dead weight (I'll call it $p_t$). If I drop it at $h/2$ the energy I'll need to reach orbit is
 $$W_{min2} = \frac{h}{2}(P_R + p_e) + \frac{h}{2}(P_R - p_t) + W'_2 = hP_R + W'_2 -  \frac{h}{2}(p_t - p_e).$$
If we sightly reduce the amount of initial fuel we can make $W'_2\leq W'$. Therefore, $W_{min2}$ is clearly less than $W_{min}$ as long as the engine you had to add is lighter than the tank you dropped.
